Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 82 / 1

The limiting polynomial has a double root at $x=-1$, so two roots require a Puiseux series. For positive $ϵ$, put $\delta =\sqrt{ϵ}$ and $x=-1+a\delta +b{\delta }^2+\cdots$. Substitution and dominant balance giveThus $a=\pm 1$ and $b=-3/2$. The two finite roots have the three-term asymptotic expansionsThis is square-root splitting of a double polynomial root. For negative $ϵ$, the same formulas apply with a chosen complex square root and describe a complex conjugate pair.

The third root is a divergent algebraic root in a singular perturbation. Set $x=X/ϵ$: the equation for $X$ is $X^3+X^2+2ϵX+{ϵ}^2=0$. Its nonzero limiting root is $X=-1$. Substituting $X=-1+Aϵ+B{ϵ}^2+\cdots$ yields $A=2$ and $B=3$, henceAs an independent check, Vieta formulas give the sum of roots as $-1/ϵ$: the two finite-root expansions sum to $-2-3ϵ+O({ϵ}^2)$, giving the same large-root expansion.

For fixed $x>1$, write $a=\sqrt{x^2-1}$ and $A=x+a$. The maximum of the integrand occurs at $\theta =0$, andHere is the required endpoint Laplace method derivation. Outside a fixed neighborhood of zero, the integrand is exponentially smaller than $A^n$. Inside that neighborhood, use $\theta =(A/(na){)}^{1/2}u$. The quadratic term gives $e^{-u^2/2}$ and the higher terms vanish in the scaled limit. The Gaussian integral on the half-line is $\sqrt{\pi /2}$, soThis is the Hyperbolic Debye asymptotic for Legendre polynomials. At the endpoint the integrand is identically one, giving $P_n(1)=1$ exactly.

The localization width in the Laplace method is $O((na{)}^{-1/2})$. It becomes order one when $n\sqrt{x-1}=O(1)$, so the Legendre boundary layer at x equals one has $q=2$. With $x=1+\nu /n^2$ and fixed $\nu \ge 0$,uniformly in $\theta$. The boundary-layer answer is thereforeThis real integral representation of the modified Bessel function I0 gives one at $\nu =0$, agreeing with the endpoint value. Applying the same endpoint Laplace method to the integral for $I_0(z)$ gives $I_0(z)\sim e^z/\sqrt{2\pi z}$. Thus, as $\nu \to \infty$, the inner result is $e^{\sqrt{2\nu }}/\sqrt{2\pi \sqrt{2\nu }}$. The outer result has precisely this limit in the overlap $1\ll \nu \ll n$. This proves both matches rather than extending the fixed-$x$ formula directly to $x=1$.

For the derivatives, the power series $I_0(\sqrt{2\nu })=1+\nu /2+{\nu }^2/16+\cdots$ suggests their leading orders. To justify differentiation, expand the original integral itself near $x=1$. Its first three even cosine moments giveOdd cosine moments vanish, and higher even moments start at cubic order. Expanding at $x=1$ yieldsFor large $n$, the graph starts at $(1,1)$ with a very large positive slope and positive curvature. It rises through the $n^{-2}$ boundary layer and then follows the rapidly increasing outer asymptotic expansion. The sketch below shows both scales.